25.6 Convergence in Measure

Proof. It is sufficient to check the conditions of Proposition 6.1.8:

1. (FB1)

   For each $\eps, \eps', \delta, \delta' > 0$,

   \[U(\delta \wedge \delta', \eps \wedge \eps') \subset U(\delta, \eps) \cap U(\delta', \eps')\]

   
   
2. (UB3)

   For each $\eps, \delta > 0$ and $f, g, h \in \mathcal{L}^{0}(X; Y)$,

   \[\bracs{d(f, h) > \delta}\subset \bracs{d(f, g) > \delta}\cup \bracs{d(g, h) > \delta}\]

   
   

   so $U(\delta/2, \eps/2) \circ U(\delta/2, \eps/2) \subset U(\delta, \eps)$.

$\square$

Proof. (1): Let $f, g, h \in \mathcal{L}^{0}(X; Y)$, then

1. (M)

   If $\alpha(f, g) = 0$, then by continuity from above,

   \[\mu\bracs{d(f, g) > 0}= \limv{n}\mu\bracs{d(f, g) > 1/n}\le \limv{n}\frac{1}{n}= 0\]

   
   

   so $f = g$ almost everywhere.

2. (PM3)

   For each $\eps > 0$,

   \[\bracs{d(f, h) > \eps}\subset \bracs{d(f, g) > \eps/2}\cup \bracs{d(g, h) > \eps/2}\]

   
   

   so $\alpha(f, h) \le \alpha(f, g) + \alpha(g, h)$.

so $\alpha$ is a metric on $\mathcal{L}^{0}(X; Y)$, modulo almost everywhere equality.

(2): Let $f, g \in \mathcal{L}^{0}(X; Y)$. For any $\eps, \delta > 0$, if $\alpha(f, g) < \eps \wedge \delta$, then there exists $r \in (0, \eps \wedge \delta]$ such that $\mu\bracs{d(f, g) > r}\le r$. Thus

On the other hand, if $\mu\bracs{d(f, g) > \eps}\le \eps$, then $d(f, g) \le \eps$. Therefore $\alpha$ induces the uniform structure of convergence in measure.$\square$

Proof. It is sufficient to check the conditions of Proposition 6.1.8:

1. (FB1)

   For each $\eps, \eps', \delta, \delta' > 0$ and $A, A' \in \cm$ with $\mu(A), \mu(A') < \infty$,

   \[U(A \cup A', \delta \wedge \delta', \eps \wedge \eps') \subset U(A, \delta, \eps) \cap U(A', \delta', \eps')\]

   
   
2. (UB3)

   For each $\eps, \delta > 0$, $A \in \cm$ with $\mu(A) < \infty$, and $f, g, h \in \mathcal{L}^{0}(X; Y)$,

   \[\bracs{d(f, h) > \delta}\subset \bracs{d(f, g) > \delta}\cup \bracs{d(g, h) > \delta}\]

   
   

   so $U(A, \delta/2, \eps/2) \circ U(A, \delta/2, \eps/2) \subset U(A, \delta, \eps)$.

$\square$

Proof. (L) + (T) $\Rightarrow$ (In Measure): Let $\eps, \delta > 0$. By (T) then there exists $F_{1} \in \fF$ and $A \in \cm$ with $\mu(A) < \infty$ such that

By (L), there exists $F_{2} \in \fF$ with $F_{2} \subset F_{1}$ such that

Therefore

$\square$

Proof. Let $\eps > 0$, then for almost every $x \in X$, there exists $N \in \natp$ such that $d(f_{n}(x), f(x)) < \eps$ for all $n \ge N$, so

By continuity from above (Proposition 22.1.5),

$\square$

Proof, [[Theorem 2.30, Fol99]. ] (1): Since $\seq{f_n}$ is Cauchy in measure, there exists a subsequence $\seq{n_k}$ such that for each $k \in \natp$, $\mu(\bracsn{d(f_{n_k}, f_{n_{k+1}}) > 2^{-k}}) \le 2^{-k}$.

In this case, for any $K \in \natp$ and $j \ge k \ge K$,

By monotonicity and subadditivity,

so

By the First Borel-Cantelli Lemma,

Thus, for almost every $x \in X$, there exists $K \in \natp$ such that $d(f_{n_j}(x), f_{n_k}(x)) < 2^{-K+1}$ for all $j, k \ge K$. Therefore $\seq{f_n(x)}$ is Cauchy for almost every $x$, and converges almost everywhere to a Borel measurable function $f \in L^{0}(X; Y)$.

Finally, for each $K \in \natp$,

so $f_{n_k}\to f$ in measure as well.

(2): Since the uniform structure of convergence in measure on $L^{0}(X; Y)$ is defined by the Ky Fan metric, completeness follows from (1) and Proposition 8.1.4.$\square$

Proof. By Definition 26.2.2, $\int f_{\alpha} d\mu \le \int f d\mu$ for each $\alpha \in A$.

On the other hand, using Lemma 26.2.3, it is sufficient to show that

for any $\phi \in \Sigma^{+}(X, \cm)$ satisfying:

1. (i)

   There exists $\delta > 0$ such that $\phi + \delta \le f$ on $\bracs{\phi > 0}$.

2. (ii)

   $\phi \in L^{1}(X, \cm)$.

To this end, let $\eps > 0$. Since $\mu\bracs{\phi > 0}< \infty$, by (b), there exists $\alpha \in A$ such that

In which case, by linearity,

As the above holds for all $\eps > 0$, $\int \phi d\mu \le \sup_{\alpha \in A}\int f_{\alpha} d\mu$. Therefore

$\square$